My overarching question is about differentiating when you have these inverse trig functions, but listed below is the specific question I am trying to solve. If you help me with the problem, it'll help me (and others) apply it to similar questions.

Problem: y = arcsin(x) - sqrt(1 - x^2) Find dy/dx

Answer Choices: 1/2sqrt(1-x^2) or 2/sqrt(1-x^2) or (1+x)/sqrt(1-x^2) or (x^2)/sqrt(1-x^2) or 1/sqrt(1+x)

The arcsin(x) is primarily what is getting me stuck. To try to solve the problem I moved the root to the other side by adding it to both sides.

  1. y + sqrt(1 - x^2) = arcsin(x)

Then I converted the equation into a sin equation...I don't feel like this is correct

  1. sin(y + sqrt(1 - x^2)) = x

From here, if I take dy/dx of both sides, it seems utterly confusing and on the wrong track. (I believe I applied chain rule correctly, but I could be wrong)

  1. cos(y + sqrt(1 - x^2)) * [dy/dx + 1/2sqrt(1-x^2) * -2x] = 1

I also examined the square root in the problem carefully because I noticed it had a striking resemblance to another problem I saw earlier in a book:

Differentiate y = arcsin(x) 1. sin(y) = x 2. cos(y) dy/dx = 1 3. dy/dx = 1/cos(y) = 1/sqrt(1- (sin(y))^2) = 1/sqrt(1-x^2) because of the trig identity (sin(x))^2 + (cos(x))^2 = 1 & because subtracting (sin(y))^2 is the same as subtracting x^2 because of Step One conversion

So because I saw the sqrt(1-x^2) in the tougher problem I'm doing right now, I tried to find a way to utilize the technique from the earlier one, but I couldn't. So perhaps that could be the key to solving it.

Thanks in advance for your help, I appreciate it.

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    $\begingroup$ Your work up to "Differentiate ..." is unnecessary. Trying to turn this into an implicit differentiation problem just makes it harder. Here's what you should do: $$y = \arcsin(x) - \sqrt{1 - x^2} \implies y' = (\arcsin(x))' - \left(\sqrt{1-x^2}\right)'$$ You figured out $(\arcsin(x))'$ correctly (you used implicit differentiation for this part but that's all the i.d. you need in this). For $ \left(\sqrt{1-x^2}\right)'$ you use the chain rule. $\endgroup$ – user137731 Jul 20 '16 at 3:47
  • $\begingroup$ Your work up to step 3 is correct, just solve for $\frac{dy}{dx}$, and then use the angle addition formula on the cosine expression if necessary. $\endgroup$ – florence Jul 20 '16 at 3:48
  • $\begingroup$ @Bye_World Thanks, that makes sense. The workbook says to use implicit differentiation though, so that's the only reason I asked that $\endgroup$ – Kartik Chugh Jul 20 '16 at 3:53
  • $\begingroup$ @florence So if I kept going, I would get the answer? $\endgroup$ – Kartik Chugh Jul 20 '16 at 3:54
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    $\begingroup$ They must have meant for you to use implicit differentiation on the $\arcsin$ portion only because otherwise you're just causing yourself a lot of undue difficulty. $\endgroup$ – user137731 Jul 20 '16 at 3:54
  1. y + sqrt(1 - x^2) = arcsin(x)
  2. sin(y + sqrt(1 - x^2)) = x
  3. cos(y + sqrt(1 - x^2)) * [dy/dx + 1/2sqrt(1-x^2) * -2x] = 1
  4. cos(arcsin(x)) * [dy/dx - x/sqrt(1-x^2)] = 1
  5. sqrt(1-x^2) * [dy/dx - x/sqrt(1-x^2)] = 1
  6. dy/dx = 1/sqrt(1-x^2) + x/sqrt(1-x^2)
  7. dy/dx = (1+x)/sqrt(1-x^2)

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